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andrew11 [14]
2 years ago
12

A farm had 480 acres more wheat than corn. After the farmers collected 80% of the wheat and 25% of the corn, the area of the whe

at was 300 acres less than the area of the corn. What was the area of the wheat field and what was the area of the corn field?
Mathematics
1 answer:
Sergio [31]2 years ago
7 0
The area covered by corn is x
The area covered by wheat is (x+480)
After 80% of wheat was collected, the new area was:
20/10(x+480)
=1/5(x+480)
After 25% of the corn was collected, the new area was:
75/100x
=3/4x
New area equation will be:
(3/4)x=1/5(x+480)+300
(3/4x)=(1/5)x+96+300
solving for x we get:
11/20x=396
x=720
thus the original area of corn was 720 acres
Original area of wheat was 720+480=1200 acres
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Answer:

See Below

Step-by-step explanation:

The function is a piecewise function defined as:

N(t)=\left \{ {{25t+150} \ 0 \leq t \leq 6 \atop {\frac{200+80t}{2+0.05t} \ t \geq 8}} \right.

a)

We need to find the limit of the function as t goes to infinity. This means what is the max value of fish in the pond given times goes to infinity (on an on).

We will take the 2nd part of the equation since t falls into that range, t is infinity, which is definitely greater than 8.

\lim_{t \to \infty} \frac{200+80t}{2+0.05t} \\\lim_{n \to \infty} \frac{80t}{0.05t}\\ \lim_{n \to \infty} \frac{80}{0.05} =1600

This means the maximum number of fish at this pond is 1600, no matter how long it goes on.

b)

A function is continuous at a point if we have the limit and the functional value at that point same.

Functional value at t = 8 is (we use 2nd part of equation):

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We do have a value and limit also goes to this as t approaches 8.

So, function is continuous at t = 8

c)

We want to find is there a "time" when the number of fishes in the pond is 250, during t from 0 to 6. We plug in 250 into N(t) and try to find t. Make sure to use the 1st part of the piece-wise function. Shown below:

N(t)=25t+150\\250=25t+150\\25t=250-150\\25t=100\\t=4

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